What is the molarity of hydrochloric acid in a solution that contains a $3.65 \cdot g$ $3.65g$ mass of $H C l (g)$ $HCl(g)$ in $500 \cdot m L$ $500mL$ aqueous solution?

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What is the molarity of hydrochloric acid in a solution that contains a $3.65 \cdot g$ $3.65g$ mass of $H C l (g)$ $HCl(g)$ in $500 \cdot m L$ $500mL$ aqueous solution?

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Answer 1 · 2017-05-09T06:33:23

$Molarity \equiv \frac{Moles of solute}{Volume of solution}$ $"Molarity"-="Moles of solute"/"Volume of solution"$

Explanation:

And here, $molarity = \frac{\frac{3.65 \cdot g}{36.5 \cdot g \cdot m o l^{- 1}}}{500 \cdot m L \times 10^{- 3} \cdot L \cdot m L^{- 1}} = 0.200 \cdot m o l \cdot L^{- 1}$ $"molarity"=((3.65*g)/(36.5*g*mol^-1))/(500*mLxx10^-3*L*mL^-1)=0.200*mol*L^-1$ .

Is the calculation consistent dimensionally?

And thus if it is $0.200 \cdot m o l \cdot L^{- 1}$ $0.200*mol*L^-1$ with respect to hydrogen chloride, it is $0.200 \cdot m o l \cdot L^{- 1}$ $0.200*mol*L^-1$ with respect to $H_{3} O^{+}$ $H_3O^+$ , AND $C l^{-}$ $Cl^-$ .

What is the $p H$ $pH$ of this solution?

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What is the molarity of hydrochloric acid in a solution that contains a $3.65 \cdot g$ $3.65g$ mass of $H C l (g)$ $HCl(g)$ in $500 \cdot m L$ $500mL$ aqueous solution?

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Explanation:

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What is the molarity of hydrochloric acid in a solution that contains a 3.65⋅g3.65*g mass of HCl(g)HCl(g) in 500⋅mL500*mL aqueous solution?

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What is the molarity of hydrochloric acid in a solution that contains a $3.65 \cdot g$ $3.65g$ mass of $H C l (g)$ $HCl(g)$ in $500 \cdot m L$ $500mL$ aqueous solution?